Question

what is the measure of angle abd in trapezoid abcd? 24° 40° 64° 92°

Explanation:

Step1: Use triangle angle sum property

In \(\triangle ABD\), we know that the sum of angles in a triangle is \(180^{\circ}\). Let \(\angle ABD = x\).

Step2: Find the measure of \(\angle ADB\)

Since \(BC\parallel AD\) and \(BC = AD\), \(AB = CD\), trapezoid \(ABCD\) is isosceles. \(\angle CBD=\angle ADB = 24^{\circ}\) (alternate - interior angles for \(BC\parallel AD\) and transversal \(BD\)).

Step3: Calculate \(\angle ABD\)

We know that in \(\triangle ABD\), \(\angle A+\angle ABD+\angle ADB = 180^{\circ}\). Given \(\angle A=116^{\circ}\), \(\angle ADB = 24^{\circ}\). Then \(x+116^{\circ}+24^{\circ}=180^{\circ}\). So \(x=180^{\circ}-(116^{\circ} + 24^{\circ})=40^{\circ}\)

Answer:

\(40^{\circ}\)